Question: Is every finite simple group a quotient of a braid group?

Context: The braid group on two strands $ B_2 $ is isomorphic to $ \mathbb{Z} $ and so the infinite family of abelian finite simple groups (cyclic of prime order) are all quotients of $ B_2 $.

The braid group on three strands $ B_3 $ is a universal central extension of $ PSL(2, \mathbb{Z}) $, $ B_3 \cong \mathbb{Z} : PSL(2,\mathbb{Z}) $. So the infinite family of finite simple groups $ PSL(2,p)\cong PSp(2,p) $ are all quotients of $ B_3 $.

This pattern continues with higher $ B_n,n \geq 3 $. There is a quotient of $ B_n $ which is a (finite index) subgroup of $ Sp(2m,\mathbb{Z}) $ for some $ m $ ($ m $ is the rank of the homology of a certain space, details given in this paper https://link.springer.com/article/10.1007/BF02566275.) And moreover if we take this subgroup mod $ p $ then we get all of $ Sp(2m,p) $. Thus we have that $ Sp(2m,p) $ for all primes $ p $ are quotients of that $ B_n $. Assuming that every integer is the $ m $ corresponding to some $ n $ then we have that every finite simple group $ PSp(2m,p) $ is a quotient of some Braid group $ B_n $.

Since $ PSp(2m,p) \cong PSL(2,p) $ then $ PSL(2,5) $ and $ PSL(2,7) $ are indeed quotients of $ B_3 $ and all cyclic groups are quotients of $ B_2 $. The next group to check would be $ A_6 $.

So one way to answer this question would just be to show that $ A_6 $ is not a quotient of any braid group $ B_n $.

  • 1
    $\begingroup$ Since $B_3$ is a central extension of $\mathrm{PSL}(2,\mathbb Z) \cong C_2 * C_3$, a group with trivial centre is a quotient of $B_3$ iff it is $(2,3)$-generated. This includes the vast majority of finite simple groups (though not $A_6$). See mathoverflow.net/questions/365374/…. $\endgroup$ Jun 28, 2023 at 10:56
  • $\begingroup$ Also, since the pure braid group $PB_3$ (the kernel of the natural map $B_3 \to S_3$) is isomorphic to $F_2 \times \mathbb Z$, and since every finite simple group is $2$-generated, every finite simple group appears as a quotient of $PB_3$ and therefore as a section of $B_3$. $\endgroup$ Jun 28, 2023 at 12:34

1 Answer 1


The alternating group $A_6$ is not a quotient of any braid group $B_n$. This follows from direct calculation. Since $$B_n = \langle x_1, \dots, x_{n-1} \mid [x_i,x_j] = 1 ~ \text{for} ~ |i-j|>1, x_ix_{i+1}x_i = x_{i+1}x_ix_{i+1} ~\text{for}~ 1 \le i \le n-2\rangle,$$ we can enumerate homomorphisms $B_n \to A_6$ by counting $x_1,\dots,x_{n-1} \in A_6$ satisfying the braid relations. It turns out that $$|\mathrm{Hom}(B_n, A_6)| = \begin{cases}3960 &:n = 3, \\ 5400 &: n=4, \\ 360 &: n = 5. \end{cases}$$ The image of a homomorphism $B_n \to A_6$ ($n \le 5$) never has order greater than $60$. Now observe that there are exactly $|G| = 360$ homomorphisms $B_n \to A_6$ factoring as $B_n \to \mathbb Z \to A_6$. The calculation for $n=5$ demonstrates that every homomorphism $B_5 \to A_6$ has this form. Since $B_{n+1}$ is generated by two copies of $B_n$, it follows by induction that every homomorphism $B_n \to A_6$ for $n \ge 5$ factors through $\mathbb Z$ and hence generates a subgroup of order at most $5$.

  • $\begingroup$ The same approach works for $A_7$, but you have to calculate $|\mathrm{Hom}(B_6, A_7)|$. $\endgroup$ Jun 28, 2023 at 13:12
  • 2
    $\begingroup$ Same approach works for $A_8$. On the other hand I believe (but I'm not 100% sure) that $A_n$ is $(2,3)$-generated for all $n > 8$, which implies it is a quotient of $B_3$. Therefore $A_6, A_7, A_8$ are the unique alternating groups which are not quotients of braid groups. $\endgroup$ Jun 28, 2023 at 14:19

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